Re: photons
- From: "Dr ***" <paulpsremove@xxxxxxxxxx>
- Date: Fri, 15 Apr 2005 08:34:12 +0100
"Dr ***" <paulpsremove@xxxxxxxxxx> wrote in message
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"Sue..." <suzysewnshow@xxxxxxxxxxxx> wrote in message
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snip
Look at the curve between x = 1 and x = 10.
You are saying you believe there exist a way to confirm
the two curves exactly converge in real space as indicated
dr
Maybe, I am trying to work out how this was plotted, is the y axis the
input impedance to the dipole ? I have know downloaded the text to go
with the plot and have a little more idea what it was actually describing
so after studying it I may revue what I have written below if needed.
s
The x axis is distance in wave lengths. 30MHz would be 10 meters Eh?
But you know full well any matter you put into that region
to make the measurements will be a new interacting structure
behaving just as the source radiating structure.
dr
I have spent a bit more time on the, well I would if I could find my printed
copy but as I cant it so I'll just use the electronic one without my
notes:-) so its from memory which can be unreliable:-)
a) I was not able to confirm to my satisfaction that the impedance in the
far field was ~400 ohms for either or both.
b) but I was satisfied that the electric field followed a different path
through vacuum space time, in the far field anyway and due to 1/r^2 as
against 1/r^3 and that in itself suggested that the plot would not converge
at ~400 for both.
c) There seemed no explanation for this data mismatch ???? so unless you
have an explanation I will just have to work on it until it becomes clear.
d) If you pick up an electric field with an antenna the impedance is set by
the circuitry that the antenna is connected to? so how would you know that
the field at the point of pick up was ~ 400 ohms?
s
How do you know at x = 1/ 2pi that the curves actually
reverse their y axis trend ?
dr
I dont and I suspect they dont ???
Dr Ardvark
How do you manage to separate the impedance's for the electric and
magnetic
fields?
the electric field will be on a slightly different vector to the
magnetic
field due to the mag. field curveture.
s
You take the sample with either a Faraday shielded loop or a short
electrostatic probe. (A direction finding antenna incorporates both
types to resolve the 0 / 180 degree ambiguity of a loop antenna. )
dr
Wont that find the field strength, perhaps your example below will say how
to convert field strength into impedance maybe its easy obvious ? cant
remember.
s
We don't have to invent virtual spinning entities in space
if they, in reality they continue toward y = o and y = infinity
respectively
Yes that seems resonable but I think that the graph would be more
informative if the mag. field tracked out at for example a 1000Z's and
the
elec. feild tracked out at 100Z's for example although I cant think at
this
moment how to measure them.
s
Perhaps the sampling technique above makes it clearer.
d
4pi*10E-7 H mE-1 seems the calculation to find the mag. track out and
8.854*10e-12 F mE-1 for the electric field track out but I cant think
of how
to convert them into impedences at the moment so they could be applied
to
your graph.
Plenty of examples here:
http://www.google.com/search?hl=en&q=%22free+space+impedance%22&btnG=Google+
Search
Sue...
.
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